TITLE: Free groups are CT-groups
QUESTION [6 upvotes]: A group $G$ is called CT-group if being commutative elements is transitive relation on $G\setminus\{1\}$  i.e. if $ 1 \neq x,y,z\in G $ and $[x,y]=1, [y,z]=1 $ then $[x,z]=1$.
I encountered the fact that free groups are CT-group, but every proof to this is very long and work for much general case (I know its true also for hyperbolic torsion free groups, which is nice but overkill for me).
Any one knows a self contained proof for this? The papers I saw uses arguments from algebraic topology, which are way over my knowledge, I only need the case of free groups.

REPLY [8 votes]: For a completely self-contained proof see Theorem 1.8.28 in my book https://math.vanderbilt.edu/sapirmv/book/b2.pdf. The idea is that you first prove that if $uv=vu$ in the free semigroup then $u,v$ are powers of some other word $w$ which is trivially proved by induction (Theorem  1.2.9). And then prove a similar fact for free group by looking at possible cancelation in $uv$ or $vu$. Case 1 is when $u$ is not cyclically reduced and case 2 is when $u$ is cyclically reduced.